TITLE: How do I prove that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite?
QUESTION [21 upvotes]: I need help proving that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite. The hint that came with the problem is: Consider the numbers $2+(n+1)!,3+(n+1)!,...,n+(n+1)!,n+1+(n+1)!$.
I honestly have no idea how to begin here. I am specifically confused about what it means when it says "there exist $n$ consecutive positive integers". Does this mean that if $n = 5$, for example, then somewhere in the positive integers there are 5 consecutive composite integers, and that we want to prove that? I get that composite means that they are not prime, so they have more factors than 1 and themselves.
Any help would be greatly appreciated.
Thank you!

REPLY [2 votes]: proof:
Let n be any natural number and consider the numbers 2 + (n + 1)!, 3 + (n + 1)!, 4 + (n + 1)!, ....., n + (n + 1)!, (n + 1) + (n + 1)!. 
observe that 2 divides the number 2 + (n + 1)!, 3 divives the number 3 +  (n + 1)!
and n divides the n + (n + 1)! likewise (n + 1) divides (n + 1)!. In short i divides i + (n + 1)! where i is an element of the set. {2, 3, 4, ..., n, n + 1} the reason been you can always factor out i. 
Likewise the numbers 2 + (n + 1)!, 3 + (n + 1)!, 4 + (n + 1)!, ....., n + (n + 1)!, (n + 1) + (n + 1)! are all consecutive and their count is n. QED